Leetcode dp Interleaving String

Interleaving String

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Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

For example,

Given:

s1 = "aabcc",

s2 = "dbbca",

When s3 = "aadbbcbcac",
return true.

When s3 = "aadbbbaccc", return false.

题意：判断某一个字符串是否是由另外两个字符串交织构成的

思路：dp

设dp[i][j]表示s3[0...i+j)能否由s1[0...i)和s2[0...j)交织构成

dp[i][j] = dp[i-1][j] ,if s3[i + j - 1] == s1[i - 1]

dp[i][j] = dp[i][j-1] ,if s3[i + j - 1] == s2[j - 1]

dp[i][j] = false, otherwise

初始化 dp[i][j]

dp[0][k] = true , if s2[0...k) == s3[0...k), 0 <= k <= j

dp[0][k] = false, otherwise

dp[k][0] = true , if s1[0...k) == s3[0...k), 0 <= k <= i

dp[k][0] = false, otherwise

复杂度：时间O(n^2)，空间O(n^2)

bool isInterleave(string s1, string s2, string s3){
	int len1 = s1.size(), len2 = s2.size(), len3 = s3.size();
	if(len1 + len2 != len3) return false;
	vector<vector<bool> > dp(len1 + 1, vector<bool>(len2 + 1, false));
	//初始化
	dp[0][0] = true;
	for(int i = 1; i <= len1; ++i){
		dp[i][0] = dp[i - 1][0] && (s1[i - 1] == s3[i - 1]);
	}
	for(int j = 1; j <= len2; ++j){
		dp[0][j] = dp[0][j - 1] && (s2[j - 1] == s3[j - 1]);
	}
	//迭代
	for(int i = 1; i <= len1; ++i){
		for(int j = 1; j <= len2; ++j){
			dp[i][j] = (s3[i + j - 1] == s1[i - 1] && dp[i - 1][j]  ||   s3[i + j - 1] == s2[j - 1] && dp[i][j - 1]);
		}
	}
	return dp[len1][len2];
}